Question

A small immersion heater is rated at 375W. The specific heat of water is 4186 J/kg?C?.

Estimate how long it will take to heat a cup of soup (assume this is 250 mL of water) from 15?C to 64?C. Ignore the heat loss to the surrounding environment.

Answer 1

energy = P t
energy = m c delta(T)
250 mL is 250 g
set them equal and solve for t
375 t = 250(4.186)(49.0)

I get about 137 sec or 2 min 17 sec

Answer 2

Answer 3

SOLUTION :

Mass of 250 mL water = 250 g = 0.25 kg

Heat required to heat this water to 64ºC from 15ºC

= m s (t2 - t1)

= 0.25 * 4186 * (64 - 15) 

= 51278.5 J 

Let time taken is t sec by the 375 W immersion heater .

Electrical energy consumed

 = 345 * t      watt-sec 

= 375 t        J ( since 1 watt-sec = 1 J)

As per conservation of energy :

Electrical energy consumed = Heat energy generated to heat the water

=> 375 t = 51278.5 

=> t = 51278.5 / 335 = 136.74 sec = 2 min. 16.74 secs.

Time needed to heat the water as required = 136.74 secs = 2 minutes 16.74 secs (ANSWER).